0.00/0.45	YES
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	  eval#(I0, I1) -> eval#(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0]
0.00/0.45	  eval#(I2, I3) -> eval#(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3]
0.00/0.45	  eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]
0.00/0.45	R = 
0.00/0.45	  eval(x, y) -> eval(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	  eval(I0, I1) -> eval(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0]
0.00/0.45	  eval(I2, I3) -> eval(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3]
0.00/0.45	  eval(I4, I5) -> eval(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]
0.00/0.45	
0.00/0.45	The dependency graph for this problem is:
0.00/0.45	  0 -> 0, 3
0.00/0.45	  1 -> 
0.00/0.45	  2 -> 
0.00/0.45	  3 -> 0, 3
0.00/0.45	Where:
0.00/0.45	  0) eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	  1) eval#(I0, I1) -> eval#(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0]
0.00/0.45	  2) eval#(I2, I3) -> eval#(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3]
0.00/0.45	  3) eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]
0.00/0.45	
0.00/0.45	We have the following SCCs.
0.00/0.45	  { 0, 3 }
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	  eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]
0.00/0.45	R = 
0.00/0.45	  eval(x, y) -> eval(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	  eval(I0, I1) -> eval(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0]
0.00/0.45	  eval(I2, I3) -> eval(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3]
0.00/0.45	  eval(I4, I5) -> eval(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]
0.00/0.45	
0.00/0.45	We use the reverse value criterion with the projection function NU:
0.00/0.45	  NU[eval#(z1,z2)] = z1
0.00/0.45	
0.00/0.45	This gives the following inequalities:
0.00/0.45	  y > x && x > 0 && y > 0 && y >= x  ==>  x >= x
0.00/0.45	  I4 > I5 && I4 > 0 && I5 > 0  ==>  I4 > I4 - I5  with  I4 >= 0
0.00/0.45	
0.00/0.45	We remove all the strictly oriented dependency pairs.
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	R = 
0.00/0.45	  eval(x, y) -> eval(x, y - x) [y > x && x > 0 && y > 0 && y >= x]
0.00/0.45	  eval(I0, I1) -> eval(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0]
0.00/0.45	  eval(I2, I3) -> eval(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3]
0.00/0.45	  eval(I4, I5) -> eval(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0]
0.00/0.45	
0.00/0.45	We use the reverse value criterion with the projection function NU:
0.00/0.45	  NU[eval#(z1,z2)] = z2 + -1 * z1
0.00/0.45	
0.00/0.45	This gives the following inequalities:
0.00/0.45	  y > x && x > 0 && y > 0 && y >= x  ==>  y + -1 * x > y - x + -1 * x  with  y + -1 * x >= 0
0.00/0.45	
0.00/0.45	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.43	EOF